Adjoint representation, standard basis - $\mathfrak{sl}(2)$ and Killing form

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Let $L$ be the Lie algebra $\mathfrak{sl}_{2}$ (char ${F}$ $\neq$ 2).

Take as standard basis for $L$ the three matrices:

$x=\begin{pmatrix} 0&1\\0&0\end{pmatrix}, y= \begin{pmatrix} 0&0\\1&0\end{pmatrix}$ and $h=\begin{pmatrix} 1&0\\0&-1\end{pmatrix}$.

When using the adjoint representation, the three matrices become:

$ad_{x}=\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}, ad_{y}=\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix}$ and $ad_{h}=\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix}$.

Therefore $\kappa$ (the Killing form) has matrix

$\begin{pmatrix} 0&0&4\\0&8&0\\4&0&0\end{pmatrix}$.

Can someone explain how the adjoint and further, the Killing form, are calculated here?

I see now how to calculate the adjoint and the Killing form.

$ad_{x}(x)$ = $[xx -xx] = 0$

$ad_{x}(h) = [xh - hx]$ = $\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 1&0\\0&-1\end{pmatrix}$ - $\begin{pmatrix} 1&0\\0&-1\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}$ = $\begin{pmatrix} 0&-2\\0&0\end{pmatrix} = -2x$

$ad_{x}(y) = [xy - yx]$ = $\begin{pmatrix} 0&1\\0&0\end{pmatrix} \times \begin{pmatrix} 0&0\\1&0\end{pmatrix}$ - $\begin{pmatrix} 0&0\\-1&0\end{pmatrix}\times\begin{pmatrix} 0&1\\0&0\end{pmatrix}$ = $\begin{pmatrix} 1&0\\0&-1\end{pmatrix} = h$

Then, these become the three columns of $ad_{x}$. Do the same for $y$ and $h$.

For the Killing form,

$\kappa(x,x) = tr(ad_{x}ad_{x})$
$\kappa(x,h) = tr(ad_{x}ad_{h})$
$\kappa(x,y) = tr(ad_{x}ad_{y})$

$tr(ad_{x}ad_{x}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0$.

$tr(ad_{x}ad_{h}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 2&0&0\\0&0&0\\0&0&-2\end{pmatrix})= tr(\begin{pmatrix} 0&&\\&0&\\&&0\end{pmatrix}) = 0$.

$tr(ad_{x}ad_{y}) =tr(\begin{pmatrix} 0&-2&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix} 0&0&0\\-1&0&0\\0&2&0\end{pmatrix})= tr(\begin{pmatrix} 2&&\\&2&\\&&0\end{pmatrix}) = 4$.

Thus, these three entries are the entries for the first row.

Repeat for $h$ and $y$.